jee main examination 03-04-2016 code-e
Post on 02-Jan-2022
4 Views
Preview:
TRANSCRIPT
CAREER POINT
CAREER POINT, CP Tower, IPIA, Road No.1, Kota (Raj.), : 0744-6630500 | www.ecareerpoint.com Email: info@cpil.in Page # 1
[ CODE – E ]
JEE Main Exam 2016 (Paper & Solution)
Code – E Date : 03-04-2016
Part A – PHYSICS Q.1 A student measures the time period of 100 oscillations of a simple pendulum four times. The data set is 90 s, 91 s,
95 s and 92 s. If the minimum division in the measuring clock is 1 s, then the reported mean time should be:
(1) 92 ± 2 s (2) 92 ± 5.0 s (3) 92 ± 1.8 s (4) 92 ± 3 s
Ans. [1]
Sol. Time period of 100 oscillations
90 sec, 91 sec, 95 sec, 92 sec
Mean value of time = 4
92959190 +++
tm = 92 sec
Absolute error
|Δt1| = |tm – t1| = 2 sec
|Δt2| = |tm – t2| = 1 sec
|Δt3| = |tm – t3| = 3 sec
|Δt4| = |tm – t4| = 0 sec
mean absolute error
Δtmean = 4
0312 +++
Δtmean = 1.5 sec ≈ 2s
mean time (92 ± 2 sec)
CAREER POINT
CAREER POINT
CAREER POINT, CP Tower, IPIA, Road No.1, Kota (Raj.), : 0744-6630500 | www.ecareerpoint.com Email: info@cpil.in Page # 2
[ CODE – E ]
Q.2 A particle of mass m is moving along side of a square of side 'a', with a uniform speed v in the x-y plane as shown in the figure:
45º
a
A
Da
C
a
Ba
R
x
y
O
v
vv
v
Which of the following statements is false for the angular momentum →L about the origin ?
(1) k̂R2
mvL −=→
when the particle is moving from A to B
(2) k̂a2
RmvL ⎥⎦
⎤⎢⎣
⎡−=
→ when the particle is moving from C to D.
(3) k̂a2
RmvL ⎥⎦
⎤⎢⎣
⎡+=
→ when the particle is moving from B to C.
(4) k̂R2
mvL =→
when the particle is moving from D to A.
Ans. [2, 4] Sol. When particle move along AB
)k̂(mvrL −= ⊥
→
)k̂(2
RmgL −=→
when particle move alobng C to D
)k̂(mvrL ⊥
→=
)k̂(a2
RmvL ⎥⎦
⎤⎢⎣
⎡+=
→
when particle move from B to C
)k̂(mvrL ⊥
→= = )k̂(a
2Rmv ⎟⎟
⎠
⎞⎜⎜⎝
⎛+
when particle move from D to A
)k̂(mvrL −= ⊥
→ = )k̂(
2Rmv −
so (2) and (4)are false
CAREER POINT
CAREER POINT, CP Tower, IPIA, Road No.1, Kota (Raj.), : 0744-6630500 | www.ecareerpoint.com Email: info@cpil.in Page # 3
[ CODE – E ]
Q.3 A point particle of mass m, moves along the uniformly rough track PQR as shown in the figure. The coefficient of friction, between the particle and the rough track equals μ. The particle is released, from rest, from the point P and it comes to rest at a point R. The energies, lost by the ball, over the parts, PQ and QR, of the track, are equal to each other, and no energy is lost when particle changes direction from PQ to QR.
The values of the coefficient of friction μ and the distance x(= QR), are, respectively close to: P
RQHorizontal
Surface
30º
h = 2m
(1) 0.2 and 6.5 m (2) 0.2 and 3.5 m (3) 0.29 and 3.5 m (4) 0.29 and 6.5 m
Students may find similar question in CP exercise sheet : [JEE Main, Chapter : Work Energy & power, Exercise # 4, Page No.41, Q.32]
Ans. [3] Sol. From work-energy theorem
KWWW21 ffg Δ=++
mgh + 2W = 0 [ ]WWW21 ff ==∵
W = – 2
mgh
and 2
mghW1f −=
– μmg cos30º(4) = – 2
mgh
μ = 0.29
and 2
mghW2f +=
– μmgx = – 2
mgh
x = μ2h = 3.5 m
Q.4 A person trying to lose weight by burning fat lifts a mass of 10 kg upto a height of 1m 1000 times. Assume that the potential energy lost each time he lowers the mass is dissipated. How much fat will he use up considering the work done only when the weight is lifted up ? Fat supplies 3.8 × 107 J of energy per kg which is converted to mechanical energy with a 20% efficiency rate. Take g = 9.8 ms–2 :
(1) 2.45 × 10–3 kg (2) 6.45 × 10–3 kg (3) 9.89 × 10–3 kg (4) 12.89 × 10–3 kg Ans. [4]
CAREER POINT
CAREER POINT, CP Tower, IPIA, Road No.1, Kota (Raj.), : 0744-6630500 | www.ecareerpoint.com Email: info@cpil.in Page # 4
[ CODE – E ]
Sol. PE = 1000 × mgh = 1000 × 10 × 9.8 × 1 = 98000 J
machenical energy per unit mass = 100
2 × 3.8 × 107
= 7.6 × 105 J/kg
∴ Fat burn = kg106.7
98005×
= 12.89 × 10–3 kg Q.5 A roller is made by joining together two cones at their vertices O. It is kept on two rails AB and CD which are
placed asymmetrically (see figure), with its axis perpendicular to CD and its centre O at the centre of line joining AB and CD (see figure). It is given a light push so that it starts rolling with its centre O moving parallel to CD in the direction shown. As it moves, the roller will tend to :
B D
CA
O
(1) turn left. (2) turn right (3) go straight (4) turn left and right alternately.
CAREER POINT
CAREER POINT, CP Tower, IPIA, Road No.1, Kota (Raj.), : 0744-6630500 | www.ecareerpoint.com Email: info@cpil.in Page # 5
[ CODE – E ]
Ans. [1]
Sol.
B D
CA
O
Friction
When inner point touch
When this cone proceeds then on AB rail the inner point-will touch the rail whose velocity in forward direction which slip forward and friction acts in backward direction. Due to this friction the system turn left.
Q.6 A satellite is revolving in a circular orbit at a height 'h' from the earth's surface (radius of earth R; h << R). The minimum increases in its orbital velocity required, so that the satellite could escape from the earth's gravitational field, is close to : (Neglect the effect of atmosphere.)
(1) gR2 (2) gR (3) 2/gR (4) )12(gR −
Students may find similar question in CP exercise sheet : [JEE Main, Chapter : Gravitation, Theory #12, Page No. 12, Q.g(iii) + d(iii)] [JEE Advance, Chapter : Gravitation, Exercise # 5B, Page No. 28, Q.8]
Ans. [4] Sol. Orbital velocity of a satellite revolving in a circular orbit at a height (h << R)
v0 = Rg
and the velocity required to escape is
ve = Rg2
So increase in velocity
Δv = ve – v0 = ( 2 – 1) Rg
Q.7 A pendulum clock loses 12 s a day if the temperature is 40ºC and gains 4s a day if the temperature is 20ºC.
The temperature at which the clock will show correct time, and the co-efficient of linear expansion (α) of the metal of the pendulum shaft are respectively :
(1) 25ºC; α = 1.85 × 10–5/ºC (2) 60ºC; α = 1.85 × 10–4/ºC (3) 30ºC; α = 1.85 × 10–3/ºC (4) 55ºC; α = 1.85 × 10–2/ºC
Students may find similar question in CP exercise sheet : [JEE Main, Chapter : Thermal Expansion, Exercise # 1, Page No.61, Q.10] [JEE Advance, Chapter : Thermal Expansion, Exercise # 4, Page No.87, Q.15]
Ans. [1] Sol. Let at temperature θ, clock gives correct time
CAREER POINT
CAREER POINT, CP Tower, IPIA, Road No.1, Kota (Raj.), : 0744-6630500 | www.ecareerpoint.com Email: info@cpil.in Page # 6
[ CODE – E ]
ΔT = ⎟⎠⎞
⎜⎝⎛ θΔα
21 T, T = 1 day = 86400s
12 = 21
α (40 – θ)T ... (i)
4 = 21
α (θ – 20) T ...(ii)
4
12 = 20–40
−θθ
3θ – 60 = 40 – θ θ = 25ºC
from (ii) 4 = 21
α (25 – 20) × 86400
α = 8640058
× = 1.85 × 10–5/ºC
Q.8 An ideal gas undergoes a quasi static, reversible process in which its molar heat capacity C remains constant. If during this process the relation of pressure P and volume V is given by PVn = constant, then n is given by (Here CP and CV are molar specific heat at constant pressure and constant volume, respectively):
(1) V
P
CCn = (2)
V
P
CCCCn
−−
= (3) V
P
CCCCn
−−
= (4) P
V
CCCCn
−−
=
Students may find similar question in CP exercise sheet : [JEE Advance, Chapter : Thermodynamics, Exercise # 1, Page No. 25, Q.24]
Ans. [2]
Sol. C = CV + ndTPdV =
n1R
1R
−+
−γ
C = n1
R
1CC
R
V
P −+
−
on solving
V
P
CCCCn
−−
=
Q.9 'n' moles of an ideal gas undergoes a process A → B as shown in the figure. The maximum temperature of the gas during the process will be:
P
2P0
P0
V0 2V0 V
B
A
(1) nR4VP9 00 (2)
nR2VP3 00 (3)
nR2VP9 00 (4)
nRVP9 00
CAREER POINT
CAREER POINT, CP Tower, IPIA, Road No.1, Kota (Raj.), : 0744-6630500 | www.ecareerpoint.com Email: info@cpil.in Page # 7
[ CODE – E ]
Students may find same question in CP exercise sheet : [JEE Main, Chapter : KTG, Exercise # 4, Page No.28, Q.11] [JEE Advance, Chapter : Thermodynamics, Exercise # 6, Page No. 49, Q.17]
Ans. [1]
Sol.
P
2P0
P0
V0 2V0 V
B(2V0, P0)
A(V0, 2P0)
P – 2P0 = 00
00
VV2P2P
−− (V – V0)
P = 2P0 – 0
0
VP V + P0
P = – VVP
0
0⎟⎟⎠
⎞⎜⎜⎝
⎛ + 3P0
00
0 P3VVP
VnRT
+⎟⎟⎠
⎞⎜⎜⎝
⎛−=
T = ⎥⎦
⎤⎢⎣
⎡+− VP3V
VP
nR1
02
0
0
for Tmax, 0dVdT
=
0P3VVP2
nR1
00
0 =⎥⎦
⎤⎢⎣
⎡+−
0V23V =
∴ Tmax = ⎥⎥⎦
⎤
⎢⎢⎣
⎡×+⎟
⎠⎞
⎜⎝⎛×− 00
2
00
0 V23P3V
23
VP
nR1
= ⎥⎦⎤
⎢⎣⎡ +− 0000 VP
29VP
49
nR1 =
nR4VP9 00
CAREER POINT
CAREER POINT, CP Tower, IPIA, Road No.1, Kota (Raj.), : 0744-6630500 | www.ecareerpoint.com Email: info@cpil.in Page # 8
[ CODE – E ]
Q.10 A particle performs simple harmonic motion with amplitude A. Its speed is trebled at the instant that it is at a
distance 3A2 from equilibrium position. The new amplitude of the motion is:
(1) 413A (2) 3A (3) 3A (4)
3A7
Students may find similar question in CP exercise sheet :
[JEE Main, Chapter : SHM, Exercise # 1, Page No.25, Q.17]
[JEE Advance, Chapter : SHM, Exercise # 5A, Page No. 34, Q.10]
Ans. [4]
Sol. Velocity at any position is given by
v = ω 22 xA −
at x = 3A2
v = ω9A4A
22 − ⇒ v =
35 Aω
if speed at x = 3A2 is tripled i.e.
v′ = 3v = 5 Aω
So new amplitude A′ is
v′ = ω2
2
3A4A ⎟
⎠⎞
⎜⎝⎛−′
5 Aω = ω9A4A
22 −′
A′ = 3A7
Q.11 A uniform string of length 20m is suspended from a rigid support. A short wave pulse is introduced at its lowest end. It starts moving up the string. The time taken to reach the support is: (take g = 10 ms–2)
(1) s22π (2) 2 s (3) s22 (4) s2
Students may find similar question in CP exercise sheet :
[JEE Main, Chapter :Wave Motion, Exercise # 2, Page No.29, Q.8]
[JEE Advance, Chapter : Transverse waves, Exercise # 4, Page No.36, Q.15]
Ans. [3]
Sol. Let m → mass/length T = (xm)g
CAREER POINT
CAREER POINT, CP Tower, IPIA, Road No.1, Kota (Raj.), : 0744-6630500 | www.ecareerpoint.com Email: info@cpil.in Page # 9
[ CODE – E ]
mTV =
= m
g)xm(
= gx
a = gxdxdgx
dxVdV
=
= x2
1ggx ××
a = 2g = constant
∴ = 21 at2
t = a2 =
2/g2 =
10202 = s22
Q.12 The region between two concentric spheres of radii 'a' and 'b', respectively (see figure), has volume charge density
ρ = rA , where A is a constant and r is the distance from the centre. At the centre of the spheres is a point
charge Q. The value of A such that the electric field in the region between the spheres will be constant, is
aQ b
(1) 2a2Qπ
(2) )ab(2
Q22 −π
(3) )ba(
Q222 −π
(4) 2aQ2
π
Ans. [1]
Sol. 0
inqdS·E∈
=∫→→
E · 4πr2 = 0
r
ardVQ
∈
ρ+ ∫ =
E · 4πr2 = 0
r
a
2drr4rAQ
∈
π+ ∫
x
TV
ρ
Qa
r
b
A
CAREER POINT
CAREER POINT, CP Tower, IPIA, Road No.1, Kota (Raj.), : 0744-6630500 | www.ecareerpoint.com Email: info@cpil.in Page # 10
[ CODE – E ]
E = ⎥⎥⎦
⎤
⎢⎢⎣
⎡⎟⎟⎠
⎞⎜⎜⎝
⎛ −π+
∈π 2arA4Q
r41 22
20
E = )]ar(A2Q[r4
1 222
0−π+
∈π
E = 20
22
20 r2
)ar(Ar4
Q∈
−+
∈π = 2
0
2
02
0 r2Aa
2A
r4Q
∈−
∈+
∈π
∵ E = constant
∴ E → Independent of r
∴ 20
2
20 r2
Aar4
Q∈
=∈π
2a2QAπ
=
Q.13 A combination of capacitors is set up as shown in the figure. The magnitude of the electric field, due to a point
charge Q (having a charge equal to the sum of the charges on the 4 µF and 9 µF capacitors), at a point distant 30 m from it, would equal
4µF3µF
9µF
2µF
+ –8V
(1) 240 N/C (2) 360 N/C (3) 420 N/C (4) 480 N/C
Students may find similar question in CP exercise sheet : [JEE Main, Chapter : Capacitance, Exercise #2, Page No. 43, Q.18] [JEE Advance, Chapter : Capacitance, Exercise # 2, Page No.24, Q.9]
Ans. [3] Sol.
4µF 3µF
9µF
8V
q q2 q1
+
2µF
8V
q = CeqV = 8124124
×⎟⎠⎞
⎜⎝⎛
+× = 24 µC
CAREER POINT
CAREER POINT, CP Tower, IPIA, Road No.1, Kota (Raj.), : 0744-6630500 | www.ecareerpoint.com Email: info@cpil.in Page # 11
[ CODE – E ]
q1 = 39
9+
× 24 µC = 18 µC
∴ Q = q + q1 = 24 + 18 = 42 µC
E = 2rKQ = 2
69
)30(1042109 −××× = 420 N/C
Q.14 The temperature dependence of resistances of Cu and undoped Si in the temperature range 300-400 K, is best described by
(1) Linear increase for Cu, linear increase for Si (2) Linear increase for Cu, exponential increase for Si (3) Linear increase for Cu, exponential decrease for Si (4) Linear decrease for Cu, linear decrease for Si
Students may find similar question in CP exercise sheet : [JEE Main, Chapter : Current Electricty, Exercise #1, Page No.51, Q.15]
Ans. [3] Sol. Cu is metal, its resistance increases linearly on increasing temperature, where as Si is semiconductor which
resistance decreases exponentially on increasing temperature. Q.15 Two identical wires A and B, each of length 'l', carry the same current I. Wire A is bent into a circle of radius R and
wire B is bent to form a square of side 'a'. If BA and BB are the values of magnetic field at the centres of the circle
and square respectively, then the ratio B
A
BB is
(1) 8
2π (2) 216
2π (3) 16
2π (4) 28
2π
Students may find same question in CP exercise sheet : [JEE Main, Chapter :Magnetic effect of curernt, Exercise # 5B, Page No. 69, Q.61] [JEE Advance, Chapter : Magnetic effect of current, Exercise # 1, Page No. 32, Q.9]
Ans. [4] Sol.
B1
R
I
A
45º45º
r
a
B2
I
B
= 2πR = 4a
R = π2
a = 4
CAREER POINT
CAREER POINT, CP Tower, IPIA, Road No.1, Kota (Raj.), : 0744-6630500 | www.ecareerpoint.com Email: info@cpil.in Page # 12
[ CODE – E ]
B1 = R2Iµ0 =
π×
22
Iµ0 = Iµ0π
B2 = 4]sin[sinr4Iµ0 ×β+α
π = 4]º45sinº45[sin
2a4
Iµ0 ×+π
= aIµ22 0
π =
4
Iµ22 0π
= π
Iµ28 0
2
1
BB =
π
π
Iµ28
Iµ
0
0
= 28
2π
Q.16 Hysteresis loops for two magnetic materials A and B are given below : B
H
(A)
B
H
(B) These materials are used to make magnets for electric generators, transformer core and electromagnet core.
Then it is proper to use (1) A for electric generators and transformers (2) A for electromagnets and B for electric generators (3) A for transformers and B for electric generators (4) B for electromagnets and transformers
Ans. [4]
Sol. fig. A → Hard magnetic material
fig. Β → soft magnetic material Transformer core and electromagnets are made by soft magnetic materials, which are easy to magnetise and demagnetise where as for electric generators we use hard magnetic materials.
Q.17 An arc lamp requires a direct current of 10 A at 80 V to function. If it is connected to a 220 V (rms), 50 Hz AC
supply, the series inductor needed for it to work is close to (1) 80 H (2) 0.08 H (3) 0.044 H (4) 0.065 H
Students may find similar question in CP exercise sheet : [JEE Main, Chapter :Alternating current, Exercise # 2, Page No. 37, Q.41] [JEE Advance, Chapter : Alternating current, Exercise # 3, Page No. 36, Q.17]
Ans. [4]
CAREER POINT
CAREER POINT, CP Tower, IPIA, Road No.1, Kota (Raj.), : 0744-6630500 | www.ecareerpoint.com Email: info@cpil.in Page # 13
[ CODE – E ]
Sol.
~ 220V-50 Hz
R L
Resistance of lamp R = iV =
1080 = 8 Ω
VR = 80 iR = 80
RXR
2202L
2 ⎟⎟⎟
⎠
⎞
⎜⎜⎜
⎝
⎛
+ = 80
2L
2 XR + = 4
11 R
2L
2 XR + = 2R16121
2LX = 2R
16105 = 28
16105
×
2LX = 420
XL = 20.5 2πfL = 20.5 2 × 3.14 × 50 L = 20.5 L = 0.065 H Q.18 Arrange the following electromagnetic radiations per quantum in the order of increasing energy : A : Blue light B : Yellow light C : X-ray D : Radiowave
(1) D, B, A, C (2) A, B, D, C (3) C, A, B, D (4) B, A, D, C Students may find similar question in CP exercise sheet : [JEE Advance, Chapter : Electromagnetic Wave, Exercise # 3, Page No. 101, Q.13]
Ans. [1] Sol. D, B, A, C
Q.19 An observer looks at a distant tree of height 10 m with a telescope of magnifying power of 20. To the observer the tree appears :
(1) 10 times taller (2) 10 times nearer (3) 20 times taller (4) 20 times nearer
Students may find similar question in CP exercise sheet : [JEE Main, Chapter : Electromagnetic wave, Exercise # 3, Page No.90, Q.13]
CAREER POINT
CAREER POINT, CP Tower, IPIA, Road No.1, Kota (Raj.), : 0744-6630500 | www.ecareerpoint.com Email: info@cpil.in Page # 14
[ CODE – E ]
Ans. [4]
Sol. Telescope provides angular magnification i.e. angle formed by image is more than that of object and it forms
image closer to the instrument so angle formed by image is increased.
Hence Ans.(4)
i.e., 20 times nearer
Q.20 The box of a pin hole camera, of length L, has a hole of radius a. It is assumed that when the hole is illuminated by a
parallel beam of light of wavelength λ the spread of the spot (obtained on the opposite wall of the camera) is the
sum of its geometrical spread and the spread due to diffraction. The spot would then have its minimum size
(say bmin) when :
(1) a = L
2λ and bmin = ⎟⎟⎠
⎞⎜⎜⎝
⎛ λL
2 2 (2) a = Lλ and bmin = ⎟⎟
⎠
⎞⎜⎜⎝
⎛ λL
2 2
(3) a = Lλ and bmin = L4λ (4) a = L
2λ and bmin = L4λ
Ans. [3]
Sol. Width of CM = a
λD2
= a
λL2 = L
L2λ
λ
= L4λ
Q.21 Radiation of wavelength λ, is incident on a photocell. The fastest emitted electron has speed v. If the wavelength is
changed to 4
3λ , the speed of the fastest emitted electron will be :
(1) > v21
34
⎟⎠⎞
⎜⎝⎛ (2) < v
21
34
⎟⎠⎞
⎜⎝⎛
(3) = v21
34
⎟⎠⎞
⎜⎝⎛ (4) = v
21
43
⎟⎠⎞
⎜⎝⎛
Students may find similar question in CP exercise sheet :
[JEE Main, Chapter : Photoelectric effect, Exercise # 5B, Page No.36, Q.16]
[JEE Advance, Chapter : Photoelectric effect, Exercise # 3, Page No.24, Q.2]
Ans. [1]
CAREER POINT
CAREER POINT, CP Tower, IPIA, Road No.1, Kota (Raj.), : 0744-6630500 | www.ecareerpoint.com Email: info@cpil.in Page # 15
[ CODE – E ]
Sol. 21 mv2 =
λhc – φ …. (i)
21 m(v′)2 =
43hcλ
– φ
21 m(v′)2 =
34
λhc – φ
φ−⎟⎠⎞
⎜⎝⎛ φ+= 22 mv
21
34)'v(m
21
3
mv21
34)'v(m
21 22 φ
+⎟⎠⎞
⎜⎝⎛=
⎟⎠⎞
⎜⎝⎛> 22 mv
21
34)'v(m
21
∴ v' > v34
Q.22 Half-lives of two radioactive elements A and B are 20 minutes and 40 minutes, respectively. Initially, the samples have equal number of nuclei. After 80 minutes, the ratio of decayed numbers of A and B nuclei will be :
(1) 1 : 16 (2) 4 : 1 (3) 1 : 4 (4) 5 : 4 Students may find similar question in CP exercise sheet : [JEE Advance, Chapter : Nuclear physics & Radioactivity, Exercise # 1, Page No. 29, Q.22]
Ans. [4] Sol.
⎟⎟⎠
⎞⎜⎜⎝
⎛= min20TA
21 ⎟
⎟⎠
⎞⎜⎜⎝
⎛= .min40TB
21
N0 ↓
2N0
↓
4N0
↓
8N0
↓
16N0
No of decayed nuclei
= N0 – 16N0 =
16N15 0
N0 ↓
2N0
↓
4N0
No of decayed nuclei
= N0 – 4
N0 = 3N3 0
Ratio of decayed Nuclei = 45
4N3
16N15 00 =
CAREER POINT
CAREER POINT, CP Tower, IPIA, Road No.1, Kota (Raj.), : 0744-6630500 | www.ecareerpoint.com Email: info@cpil.in Page # 16
[ CODE – E ]
Q.23 If a, b, c, d are inputs to a gate and x is its output, then, as per the following time graph, the gate is :
d
c
b
a
x (1) NOT (2) AND (3) OR (4) NAND
Ans. [3] Sol. If one of the inputs is high then output will be high in OR gate. Q.24 Choose the correct statement :
(1) In amplitude modulation the amplitude of the high frequency carrier wave is made to vary in proportion to the amplitude of the audio signal
(2) In amplitude modulation the frequency of the high frequency carrier wave is made to vary in proportion to the amplitude of the audio signal
(3) In frequency modulation the amplitude of the high frequency carrier wave is made to vary in proportion to the amplitude of the audio signal
(4) In frequency modulation the amplitude of the high frequency carrier wave is made to vary in proportion to the frequency of the audio signal
Students may find similar question in CP exercise sheet :
[JEE Main, Chapter : Communication, Exercise # 1, Page No.101, Q.7]
Ans. [1]
Sol. (1) is correct but language of this option is not upto the mark.
Q.25 A screw gauge with a pitch of 0.5 mm and a circular scale with 50 divisions is used to measure the thickness of a thin sheet of Aluminium. Before starting the measurement, it is found that when the two jaws of the screw gauge are brought in contact, the 45th division coincides with the main scale line and that the zero of the main scale is barely visible. What is the thickness of the sheet if the main scale reading is 0.5 mm and the 25th division coincides the main scale line ?
(1) 0.75 mm (2) 0.80 mm (3) 0.70 mm (4) 0.50 mm
Students may find similar question in CP exercise sheet :
[JEE Main, Chapter : Practical physics, Page No.26, Q.19]
[JEE Advance, Chapter : Practical physics, Exercise # 3, Page No.72, Q.5]
Ans. [2]
CAREER POINT
CAREER POINT, CP Tower, IPIA, Road No.1, Kota (Raj.), : 0744-6630500 | www.ecareerpoint.com Email: info@cpil.in Page # 17
[ CODE – E ]
Sol. LC = scalecircular on div. of no.
pitch
= 50mm0.5 = 0.01 mm
Zero error e = N × LC = 5 × 0.01
= 0.05 mm
True reading = observed reading + e
= LSR + CSR + e
= 0.5 mm + 25 × 0.01 + 0.05
= 0.75 + 0.05 = 0.80 mm
Q.26 A pipe open at both ends has a fundamental frequency ƒ in air. The pipe is dipped vertically in water so that half of it is in water. The fundamental frequency of the air column is now :
(1) 2ƒ (2)
4ƒ3 (3) 2ƒ (4) ƒ
Students may find same question in CP exercise sheet : [JEE Main, Chapter :Sound wave, Exercise # 1, Page No.43, Q.26] [JEE Advance, Chapter : Sound wave, Exercise # 1, Page No. 83, Q.14]
Ans. [4] Sol.
A
N
A
= 4λ +
4λ =
2λ
λ = 2
ƒ = λv =
2v
A
N 2
2 =
4λ ⇒ λ = 2
∴ ƒ ′ = λv =
2v = ƒ
Q.27 A galvanometer having a coil resistance of 100 Ω gives a full scale deflection, when a current of 1 mA is passed through it. The value of the resistance which can convert this galvanometer into ammeter giving a full scale deflection for a current of 10 A, is :
(1) 0.01 Ω (2) 2 Ω (3) 0.1 Ω (4) 3 Ω Students may find similar question in CP exercise sheet : [JEE Main, Chapter : Magnetic effect of current, Exercise # 1, Page No.45, Q.76] [JEE Advance, Chapter : Current Electricity, Exercise # 2, Page No. 40, Q.28]
Ans. [1]
CAREER POINT
CAREER POINT, CP Tower, IPIA, Road No.1, Kota (Raj.), : 0744-6630500 | www.ecareerpoint.com Email: info@cpil.in Page # 18
[ CODE – E ]
Sol. G = 100 Ω ig = 1 mA = 10–3 A i = 10 A s = ?
s = g
g
iiGi
−
×
= 3
3
101010010
−
−
−×
= 10
10010 3 ×−
= 10–2 Ω
Q.28 In an experiment for determination of refractive index of glass of a prism by i – δ, plot, it was found that a ray incident at angle 35º, suffers a deviation of 40º and that it emerges at angle 79º. In that case which of the following is closest to the maximum possible value of the refractive index ?
(1) 1.5 (2) 1.6 (3) 1.7 (4) 1.8
Ans. [1]
Sol. i = 35º
δ = 40º
e = 79º
A = ?
δ = i + e – A
40º = 35º + 79º – A
A = 35º + 79º – 40º
A = 74º
40º
35º 79ºi
δ
μ =
2Asin
2Asin m ⎟
⎠⎞
⎜⎝⎛ +δ
CAREER POINT
CAREER POINT, CP Tower, IPIA, Road No.1, Kota (Raj.), : 0744-6630500 | www.ecareerpoint.com Email: info@cpil.in Page # 19
[ CODE – E ]
μ = º37sin
2º74sin m ⎟
⎠⎞
⎜⎝⎛ δ+
μ = 35 sin ⎟
⎠⎞
⎜⎝⎛ δ
+2
º37 m
μ < 35 sin 57º <
35 sin 60º
μ < 32
5⇒ μ < 1.44
Q.29 Identify the semiconductor devices whose characteristics are given below, in the order (a), (b), (c), (d) :
(a) V
I
(b) V
I
(c) V
I dark
illuminated
(d)
resistance
intensity of light
(1) Simple diode, Zener diode, Solar cell, Light dependent resistance (2) Zener diode, Simple diode, Light dependent resistance, Solar cell (3) Solar cell, Light dependent resistance, Zener diode, Simple diode (4) Zener diode, Solar cell, Simple diode, Light dependent resistance
Ans. [1] Sol. By theory Q.30 For a common emitter configuration, if α and β have their usual meanings, the incorrect relationship between α and
β is :
(1) α1 =
β1 + 1 (2) α =
β−β
1
(3) α = β+
β1
(4) α = 2
2
1 β+β
Students may find similar question in CP exercise sheet : [JEE Main, Chapter : Semiconductor Electronics, Theory: point no. 17, Page No. 22
Ans. [2, 4]
Sol. 111+
β=
α
or β+
β=α
1
CAREER POINT
CAREER POINT, CP Tower, IPIA, Road No.1, Kota (Raj.), : 0744-6630500 | www.ecareerpoint.com Email: info@cpil.in Page # 20
[ CODE – E ]
Part B – CHEMISTRY Q.31 At 300 K and 1 atm, 15 mL of a gaseous hydrocarbon requires 375 mL air containing 20% O2 by volume for
combustion. After combustion the gaseous occupy 330 mL. Assuming that the water formed is in liquid form and the volumes were measured at the same temperature and pressure, the formula of the hydrocarbon is :
(1) C3H6 (2) C3H8 (3) C4H8 (4) C4H10 Ans. [Bonus]
Sol. CxHy(g) + )g(O4yx 2⎟
⎠⎞
⎜⎝⎛ + ⎯→ x CO2(g) + )(OH
2y
2
Initially 15 ml 75 ml 0 – After reaction 0 0 15x – As per problem :
⎟⎠⎞
⎜⎝⎛ +
=
4yx
751
15
So, 54yx =+ ….(i)
and 15x = 330 – 300 x = 2 ….(ii) from eq. (i) & (ii) y = 12 In this problem it seems to be there is a printing error if volume of gaseous products after combustion is 345
ml instead of 330 ml. 15x = 345 – 300 x = 3
& 54yx =+
∴ y = 8 C3H8 Ans.
Q.32 Two closed bulbs of equal volume (V) containing an ideal gas initially at pressure pi and temperature T1 are
connected through a narrow tube of negligible volume as shown in the figure below. The temperature of one of the bulbs is then raised to T2. The final pressure pf is :
T1 T1pi,V pi,V
T1 T2
pf,V pf,V
(1) ⎟⎟⎠
⎞⎜⎜⎝
⎛+ 21
21i TT
TTp (2) ⎟⎟⎠
⎞⎜⎜⎝
⎛+ 21
1i TT
Tp2 (3) ⎟⎟⎠
⎞⎜⎜⎝
⎛+ 21
2i TT
Tp2 (4) ⎟⎟⎠
⎞⎜⎜⎝
⎛+ 21
21i TT
TTp2
Students may find similar question in CP exercise sheet : [JEE Main, Chapter : Gaseous State, Solved Example No. 32] [JEE Advance, Chapter : Gaseous State, Exercise. # 2, Q.39]
Ans. [3]
CAREER POINT
CAREER POINT, CP Tower, IPIA, Road No.1, Kota (Raj.), : 0744-6630500 | www.ecareerpoint.com Email: info@cpil.in Page # 21
[ CODE – E ]
Sol. Initial condition
T1 T1 pi,V pi,V
So, initial mole in both the bulb is same, say (n)
So, i
i
TRVpn =
Final condition T1 T2 pf
n1V
pf n2V
So, n1 + n2 = 2n [Total mole must be same]
⎟⎟⎠
⎞⎜⎜⎝
⎛=+
1
i
2
f
1
f
RTVP2
RTVP
RTVP
1
i
21f T
P2T1
T1P =⎥
⎦
⎤⎢⎣
⎡+
Pf = 2Pi ⎭⎬⎫
⎩⎨⎧
+ 21
2
TTT
Q.33 A stream of electrons from a heated filament was passed between two charged plates kept at a potential difference V
esu. If e and m are charge and mass of an electron, respectively, then the value of h/λ (where λ is wavelength associated with electron wave) is given by :
(1) meV (2) 2meV (3) meV (4) meV2
Students may find similar question in CP exercise sheet : [JEE Main, Chapter : Atomic Structure, Key concepts, Topic :-Wave mechanical model of atoms] Students may find same question in CP exercise sheet : [JEE Advance, Chapter : Atomic structure, Same Q., Page No.3]
Ans. [4]
Sol. ph
=λ ….(1)
p = mv ….(2)
K.E. = 2mv21
mKE2v = …..(3)
from eq. (1), (2) & (3)
mKE2m
h=λ
CAREER POINT
CAREER POINT, CP Tower, IPIA, Road No.1, Kota (Raj.), : 0744-6630500 | www.ecareerpoint.com Email: info@cpil.in Page # 22
[ CODE – E ]
mKE2h
=λ ….(4)
If electron is accelerated by potential difference of V then its K.E. is given as follows K.E. = eV ….(5) from eq. (4) and (5)
meV2h
=λ
meV2h=
λ
Q.34 The species in which the N atom is in a state of sp hybridization is :
(1) +2NO (2) −
2NO (3) −3NO (4) NO2
Students may find similar question in CP exercise sheet : [JEE Main, Chapter : Chemical bonding, Exercise. # 4, Q.26] [JEE Advance, Chapter : Chemical bonding, Exercise. # 2, Q.30]
Ans. [1]
Sol. In +2NO , 'N' has zero lone pair and two bond pairs (σ-bond). Hence, hybridization is sp.
Q.35 The heats of combustion of carbon and carbon monoxide are –393.5 and –283.5 kJ mol–1, respectively. The heat of
formation (in kJ) of carbon monoxide per mole is : (1) 110.5 (2) 676.5 (3) –676.5 (4) –110.5
Students may find similar question in CP exercise sheet : [JEE Main, Chapter : Chemical energetics, Exercise. # 4, Q.6] [JEE Advance, Chapter : Chemical energetics, Exercise. # 2, Q.12]
Ans. [4] Sol. Given
C(s) + O2(g) ⎯→ CO2(g), ΔH1 = –393.5 kJ/mol
CO(g) + )g(O21
2 ⎯→ CO2(g), ΔH2 = –283.5 kJ/mol
Object reaction,
C(s) + )g(O21
2 ⎯→ CO(g) ΔH = ?
eq (i) – eq (ii) ΔH = ΔH1 – ΔH2
= –393.5 + 283.5 = –110 KJ/mol.
Q.36 18 g glucose (C6H12O6) is added to 178.2 g water. The vapour pressure of water (in torr) for this aqueous solution
is: (1) 7.6 (2) 76.0 (3) 752.4 (4) 759.0
CAREER POINT
CAREER POINT, CP Tower, IPIA, Road No.1, Kota (Raj.), : 0744-6630500 | www.ecareerpoint.com Email: info@cpil.in Page # 23
[ CODE – E ]
Students may find similar question in CP exercise sheet : [JEE Main, Chapter : Solution & Colligative properties, Exercise. # 4, Q.16] [JEE Advance, Chapter : Solution & Colligative properties, Exercise. # 5A, Q.5]
Ans. [3] Sol. We assume the solution is prepared at boiling point temperature of water ∴ P° = 760 mm Hg
Nn
nP
PP s
+=
°−°
1.09.9
1.0760
P–760 s
+=
10
1.0760P1 s =−
1 – 0.01 = 760Ps
Ps = 752.4 mm Hg or torr. Q.37 The equilibrium constant at 298 K for a reaction A + B C + D is 100. If the initial concentration of all the four
species were 1 M each, then equilibrium concentration of D (in mol L–1) will be : (1) 0.182 (2) 0.818 (3) 1.818 (4) 1.182
Students may find similar question in CP exercise sheet : [JEE Main, Chapter : Chemical equilibrium, Exercise. # 5B, Q.5]
[JEE Advance, Chapter : Chemical equilibrium, Exercise. # 2, Q.2]
Ans. [3] Sol. A + B C + D
1 1 1 1 1–x 1–x 1+x 1+x
t = 0 eqm:
At the given state Q = 1; therefore Q < K
∴ reaction will move in forward direction.
So, 100 = )x–1)(x–1()x1)(x1( ++
10 = x–1x1 +
10 – 10x = 1 + x 11x = 9
x = 119
So, [D] = 1 + x = 1 + 119
= 1120 = 1.818 M
CAREER POINT
CAREER POINT, CP Tower, IPIA, Road No.1, Kota (Raj.), : 0744-6630500 | www.ecareerpoint.com Email: info@cpil.in Page # 24
[ CODE – E ]
Q.38 Galvanization is applying a coating of :
(1) Pb (2) Cr (3) Cu (4) Zn
Students may find similar question in CP exercise sheet :
[JEE Main, Chapter : Electro chemistry, Key concept, Topic :- Prevention from corrosion]
Ans. [4]
Sol. In Galvanization Zn coating is applied an iron.
0ZnOXE > 0
FeOXE
Q.39 Decomposition of H2O2 follows a first order reaction. In fifty minutes the concentration of H2O2 decreases from 0.5 to 0.125 M in one such decomposition. When the concentration of H2O2 reaches 0.05 M, the rate of formation of O2 will be :
(1) 6.93 × 10–2 mol min–1 (2) 6.93 × 10–4 mol min–1
(3) 2.66 L min–1 at STP (4) 1.34 × 10–2 mol min–1
Students may find similar question in CP exercise sheet :
[JEE Main, Chapter : Chemical kinetics, Exercise. # 4, Q.22]
[JEE Advance, Chapter : Chemical kinetics, Exercise. # 2, Q.43]
Ans. [2]
Sol. Here solution is given assuming unit of rate is the option to be mol L–1 min–1. For 1st order reaction,
K = 501 ln
125.05.0
K = 50
4ln min–1
Now, R = K [H2O2]1
R = ⎟⎠⎞
⎜⎝⎛
504ln × 0.05 M min–1
R = 2 × 0.693 × 10–3 mol lit–1 min–1
And, H2O2 ⎯→ H2O + 21 O2
So rate of reaction, R = ⎟⎠⎞
⎜⎝⎛
21
)OofformationofRate(R 2O2
Or 2OR =
21 R =
210693.02 3–×× mol Lit–1 min–1
2OR = 6.93 × 10–4 mol L–1 min–1
CAREER POINT
CAREER POINT, CP Tower, IPIA, Road No.1, Kota (Raj.), : 0744-6630500 | www.ecareerpoint.com Email: info@cpil.in Page # 25
[ CODE – E ]
Q.40 For a linear plot of log (x/m) versus log p in a Freundlich adsorption isotherm, which of the following statements is correct ? (k and n are constants)
(1) Both k and 1/n appear in the slope term (2) 1/n appears as the intercept (3) Only 1/n appears as the slope (4) log (1/n) appears as the intercept
Students may find similar question in CP exercise sheet : [JEE Main, Chapter : Surface chemistry, Exercise. # 1, Q.45] [JEE Advance, Chapter : Surface chemistry, Exercise. # 1, Q.8]
Ans. [3]
Sol. mx = KP1/n
log mx =
n1 log P + log K
∴ Slope is n1
Q.41 Which of the following atoms has the highest first ionization energy ?
(1) Rb (2) Na (3) K (4) Sc Ans. [4] Sol. Order of first ionization energy : Sc > Na > K > Rb Q.42 Which one of the following ores is best concentrated by froth floatation method ?
(1) Magnetite (2) Siderite (3) Galena (4) Malachite
Students may find similar question in CP exercise sheet : [JEE Main, Chapter : Metallurgy, Exercise. # 4, Q.3] [JEE Advance, Chapter : Metallurgy, Exercise. # 2, Q.43]
Ans. [3] Sol. Froth floatation method is mainly used for sulphide ores Magnetite Fe3O4 Siderite FeCO3 Galena PbS Malachite Cu(OH)2
.CuCO3 Q.43 Which one of the following statements about water in FALSE ?
(1) Water is oxidized to oxygen during photosynthesis (2) Water can act both as an acid and as a base (3) There is extensive intramolecular hydrogen bonding in the condensed phase (4) Ice formed by heavy water sinks in normal water
Ans. [3] Sol. In water, there is intermolecular hydrogen bonding, not intramolecular hydrogen bonding
CAREER POINT
CAREER POINT, CP Tower, IPIA, Road No.1, Kota (Raj.), : 0744-6630500 | www.ecareerpoint.com Email: info@cpil.in Page # 26
[ CODE – E ]
Q.44 The main oxides formed on combustion of Li, Na and K in excess of air are, respectively : (1) Li2O, Na2O and KO2 (2) LiO2, Na2O2 and K2O (3) Li2O2, Na2O2 and KO2 (4) Li2O, Na2O2 and KO2
Students may find similar question in CP exercise sheet : [JEE Main, Chapter : S-block elements, Exercise. # 1, Q.19] [JEE Advance, Chapter : S-block elements, Exercise. # 5, Q.2]
Ans. [4]
Sol. 2 Li + OLiO21
2)excess(2 ⎯→⎯Δ
2Na + 22)excess(2 ONaO ⎯→⎯Δ
K + 2)excess(2 KOO ⎯→⎯Δ
Q.45 The reaction of zinc with dilute and concentrated nitric acid, respectively, produces :
(1) N2O and NO2 (2) NO2 and NO (3) NO and N2O (4) NO2 and N2O
Students may find similar question in CP exercise sheet : [JEE Main, Chapter : p-block elements, Exercise. # 5A, Q.29] [JEE Advance, Chapter : p-block elements, Key concept, Topic :- Properties of nitric acid ]
Ans. [1]
Sol. Zn + HNO3(dil) ⎯→⎯ Zn(NO3)2 + NH4NO3 + H2O
Zn + HNO3(conc.) ⎯→⎯ Zn(NO3)2 + NO2↑ + H2O
Δ
N2O↑
Q.46 The pair in which phosphorous atoms have a formal oxidation state of +3 is :
(1) Orthophosphorous and pyrophosphorous acids (2) Pyrophosphorous and hypophosphoric acids (3) Orthophosphorous and hypophosphoric acids (4) Pyrophosphorous and pyrophosphoric acids
Students may find similar question in CP exercise sheet : [JEE Advance, Chapter : p-block elements, Key concept, Topic :- Oxyacids of phosphorus ]
Ans. [1] Sol. Orthophosphorous acid is H3PO3 and Pyrophosphorous acid is H4P2O5. In both compounds, oxidation state
of 'P' is +3 Q.47 Which of the following compounds is metallic and ferromagnetic ?
(1) TiO2 (2) CrO2 (3) VO2 (4) MnO2
Students may find similar question in CP exercise sheet :
[JEE Main, Chapter : Solid state, Key concept, Topic :- Magnetic properties of crystals]
[JEE Advance, Chapter : Solid state, Key concept, Topic :- Magnetic properties of crystals]
Ans. [2]
Sol. Ferromagnetic materials are Fe, Co, Ni, Gd, CrO2 etc.
CAREER POINT
CAREER POINT, CP Tower, IPIA, Road No.1, Kota (Raj.), : 0744-6630500 | www.ecareerpoint.com Email: info@cpil.in Page # 27
[ CODE – E ]
Q.48 The pair having the same magnetic moment is :
[At. No, : Cr = 24, Mn = 25, Fe = 26, Co = 27]
(1) [Cr(H2O)6]2+ and [CoCl4]2– (2) [Cr(H2O)6]2+ and [Fe(H2O)6]2+
(3) [Mn(H2O)6]2+ and [Cr(H2O)6]2+ (4) [CoCl4]2– and [Fe(H2O)6]2+
Students may find similar question in CP exercise sheet :
[JEE Main, Chapter : Co-ordination compounds, Solved Example No.-17]
[JEE Advance, Chapter : Co-ordination compounds, Solved Example No.-22]
Ans. [2]
Sol. [Cr(H2O)6]+2
Cr+2 = [Ar] 4s03d4
3d
↑ ↑ ↑ ↑
4s 4p 4d
sp3d2
It has 4 unpaired electrons [Fe(H2O)6]+2
Fe+2 = [Ar] 4s03d6
3d
↑↓ ↑ ↑ ↑
4s 4p 4d
sp3d2
↑
It has 4 unpaired electrons Q.49 Which one of the following complexes shows optical isomerism ?
(1) [Co(NH3)3Cl3] (2) cis[Co(en)2Cl2]Cl (3) trans[Co(en)2Cl2]Cl (4) [Co(NH3)4Cl2]Cl
Students may find similar question in CP exercise sheet : [JEE Main, Chapter : Co-ordination compounds, Exercise. # 4, Q.31] [JEE Advance, Chapter : Co-ordination compounds, Exercise. # 3, Q.26]
Ans. [2] Sol. cis-[Co(en)2Cl2]Cl shows optical isomerism. It's general case is M(aa)2b2
Cl
Co+2
en
en Cl
Cl
Co+2 enCl
enMirror
Mirror image is non-super imposable. Hence, this compound is optically active.
CAREER POINT
CAREER POINT, CP Tower, IPIA, Road No.1, Kota (Raj.), : 0744-6630500 | www.ecareerpoint.com Email: info@cpil.in Page # 28
[ CODE – E ]
Q.50 The concentration of fluoride, lead, nitrate and iron in a water sample from an underground lake was found to be 1000 ppb, 40 ppb, 100 ppm and 0.2 ppm, respectively. This water is unsuitable for drinking due to high concentration of :
(1) Fluoride (2) Lead (3) Nitrate (4) Iron Ans. [3] Sol. Substance Upper limit
Fluoride 2 ppm or 2000 ppb Lead 0.05 ppm or 50 ppb Nitrate 50 ppm Fe 0.2 ppm
In water sample, concentration of nitrate is higher than upper limit. Hence, this water is unsuitable for drinking.
Q.51 The distillation technique most suited for separating glycerol from spent-lye in the soap industry is :
(1) Simple distillation (2) Fractional distillation (3) Steam distillation (4) Distillation under reduced pressure
Students may find similar question in CP exercise sheet : [JEE Main, Chapter : Purification of organic compound, Key concept, Topic :- Distillation under reduced pressure] [JEE Advance, Chapter : Practical organic chemistry, Exercise. # 1, Q.12]
Ans. [4] Sol. Distillation under reduced pressure is the technique most suited for separation of Glycerol from Spent-lye
(Soap Solution obtain from basic hydrolysis of fats) (Mentioned in NCERRT in chapter purification methods)
Q.52 The product of the reaction given below is :
322 COK/OH.2
h/NBS.1 ⎯⎯⎯ →⎯ υ X
(1)
(2) OH
(3) O
(4) CO2H
Students may find similar question in CP exercise sheet : [JEE Advance, Chapter : Hydrocarbon , Exercise. # 3, Q.51]
Ans. [2] Sol.
Mechanismradicalfree
byationminBro
Allylic
h/NBS ⎯⎯⎯ →⎯ υ ⎯⎯⎯⎯ →← sonanceRe
OH
onSubstitutilicNucleophil
COK/OH 322 ⎯⎯⎯⎯ →⎯ Br
CAREER POINT
CAREER POINT, CP Tower, IPIA, Road No.1, Kota (Raj.), : 0744-6630500 | www.ecareerpoint.com Email: info@cpil.in Page # 29
[ CODE – E ]
Q.53 The absolute configuration of
CH3
CO2H
H OH
H Cl
is :
(1) (2R, 3S) (2) (2S, 3R) (3) (2S, 3S) (4) (2R, 3R)
Students may find similar question in CP exercise sheet : [JEE Main, Chapter : Isomerism, Exercise. # 3, Q.15] [JEE Advance, Chapter : Isomerism, Exercise. # 1, Q.36]
Ans. [2]
Sol.
CH3
C–OH
H OH
H Cl
O 1
2 S
3 R
4 2S 3R
C—Cl
C=O
H OH
OH2
3
Clockwise but lowest priority group at vertical line so configuration is (S)
CH3
C
H Cl
(2)
3
(4) (1)
Anticlockwise but lowest priority group at vertical line so configuration is (R)
4
Q.54 2-chloro-2-methylpentane on reaction with sodium methoxide in methanol yields :
(a) 3
3|
3|
252 OCH–CH
CH
CCHHC (b) 2
3
|252 CHCHCCHHC = (c) 3
3
|52 CH–CHCCHHC =
(1) All of these (2) (a) and (c) (3) (c) only (4) (a) and (b) Students may find same question in CP exercise sheet : [JEE Main, Chapter : Alkyl halide, Exercise. # 1, Q.23] [JEE Advance, Chapter : Halo alkane, Exercise. # 1, Q.23]
Ans. [1] Sol.
CH3 – CH2 – CH2 – C – Cl
CH3
CH3
OHCH
NaOCH
3
3⎯⎯⎯ →⎯
15
2 3 4CH3 – CH2 – CH = C – CH3
CH3
Elimination
2 Chloro 2 methyl Pentane +CH3 – CH2 – CH2 – C = CH2
CH3
Elimination
+
CH3 – CH2 – CH2 – C – OCH3
CH3
Substitution CH3
CAREER POINT
CAREER POINT, CP Tower, IPIA, Road No.1, Kota (Raj.), : 0744-6630500 | www.ecareerpoint.com Email: info@cpil.in Page # 30
[ CODE – E ]
Q.55 The reaction of propene with HOCl (Cl2 + H2O) proceeds through the intermediate :
(1) CH3–CH+–CH2–OH (2) CH3–CH+–CH2–Cl (3) CH3–CH(OH)– +2CH (4)CH3–CHCl– +
2CH
Students may find same question in CP exercise sheet : [JEE Main, Chapter : Hydrocarbon, Exercise. # 1, Q.54] [JEE Advance, Chapter : Hydrocarbon, Exercise. # 1, Q.45]
Ans. [2]
Sol.
CH3–CH–CH2
⊕
Cl)HOCl(
OHCl 22 ⎯⎯⎯ →⎯ + CH3–CH=CH2Propene
CH3–CH–CH2
Cl⊕
CH3–CH–CH2–Cl
OHOHΘ
Ans. is (2) CH3–CH+–CH2–Cl Q.56 In the Hofmann bromamide degradation reaction, the number of moles of NaOH and Br2 used per mole of amine
produced are : (1) One mole of NaOH and one mole of Br2. (2) Four moles of NaOH and two moles of Br2 (3) Two moles of NaOH and two moles of Br2. (4) Four moles of NaOH and one mole of Br2.
Students may find same question in CP exercise sheet : [JEE Main, Chapter : Nitrogen compound, Solved Example- 2] [JEE Advance, Chapter : Nitrogen compound, Exercise. # 4, Q.6]
Ans. [4]
Sol. C–N–H
O H
OH—
HONaMoleI
2
St⎯⎯⎯⎯⎯ →⎯
Θ
C–NH
O Θ
)MoleI(Br–Br
St
IInd Mole of NaOH
C – N – Br
O
H
–H2OΘ
C– N – Br
OΘ
Θ
⎯⎯⎯ ⎯←Br–
arrangmentRe
WOLFF N = C = O
H OH H OH
2 Moles of NaOH
NH2 + CO2 ↑ + H2O Total (4) moles of NaOH and 1 mole of Br2 is required for conversion from benzamide to per mole of amine
CAREER POINT
CAREER POINT, CP Tower, IPIA, Road No.1, Kota (Raj.), : 0744-6630500 | www.ecareerpoint.com Email: info@cpil.in Page # 31
[ CODE – E ]
Q.57 Which of the following statements about low density polythene is FALSE?
(1) Its synthesis requires high pressure.
(2) It is a poor conductor of electricity.
(3) Its synthesis requires dioxygen or a peroxide initiator as a catalyst.
(4) It is used in the manufacture of buckets, dust-bins etc.
Students may find similar question in CP exercise sheet :
[JEE Advance, Chapter : Polymer & Biomolecule, Key concept, Topic:- Addition polymer]
Ans. [4]
Sol. (1) It synthesis requires high pressure : TRUE
(2) Low density polythene is not poor conductor of electricity so it is : TRUE
(3) It synthesis requires dioxygen or a peroxide initiator as a catalyst : TRUE
(4) It is used in the manufacture of buckets, dust-bins etc : FALSE
Q.58 Thiol group is present in :
(1) Cytosine (2) Cystine (3) Cysteine (4) Methionine
Students may find similar question in CP exercise sheet :
[JEE Advance, Chapter : Polymer & Biomolecule, Key concept, Topic:- Amino acids]
Ans. [3]
Sol. Thiol (–SH) group is present in cysteine Amino Acid
Structure of Cysteine : -
H
CH2–SH
OC–HO||
NH2
Q.59 Which of the following is an anionic detergent?
(1) Sodium stearate (2) Sodium lauryl sulphate (3) Cetyltrimethyl ammonium bromide (4) Glyceryl oleate
Students may find similar question in CP exercise sheet : [JEE Main, Chapter : Polymer & Biomolecule, Key concept, Topic :- 4.9 (i)]
Ans. [2] Sol. Sodium lauryl salphate is an example of anionic detergent
CAREER POINT
CAREER POINT, CP Tower, IPIA, Road No.1, Kota (Raj.), : 0744-6630500 | www.ecareerpoint.com Email: info@cpil.in Page # 32
[ CODE – E ]
Q.60 The hottest region of Bunsen flame shown in the figure below is :
region 1
region 2region 3
region 4
(1) region 1 (2) region 2 (3) region 3 (4) region 4
Ans. [2] Sol.
region 1
region 2 (Hottest region)region 3
region 4
The very top of the inner cone is Hottest because when you have a definite inner cone, that's when there is the most air flow through the Bunsen burner. When you have a definite inner cone, that's when there is the most air flow through the Bunsen burner and the most heat and that is concentrated at the tip of inner cone.
CAREER POINT
CAREER POINT, CP Tower, IPIA, Road No.1, Kota (Raj.), : 0744-6630500 | www.ecareerpoint.com Email: info@cpil.in Page # 33
[ CODE – E ]
Part C – MATHEMATICS
Q.61 If f(x) + 2f ⎟⎠⎞
⎜⎝⎛
x1 = 3x, x ≠ 0, and S = {x ∈ R : f(x) = f(–x)}; then S :
(1) is an empty set (2) contains exactly one element (3) contains exactly two elements (4) contains more than two elements
Students may find similar question in CP exercise sheet : [JEE Advance, Chapter : Function, Ex. # 4, Q._17]
Ans. [3]
Sol. Given that f(x) + 2f ⎟⎠⎞
⎜⎝⎛
x1 = 3x, x ≠ 0 ....(1)
Replace x → x1 then 2f(x) + f ⎟
⎠⎞
⎜⎝⎛
x1 =
x3 ...(2)
from (1) & (2)
f(x) = – x + x2
∴ f(–x) = x – x2
f(x) = f(–x) (Given)
⇒ –x + x2 = x –
x2
⇒ 2x2 = 4
⇒ x2 = 2
⇒ x = ± 2
So numbers of solutions is equal to 2.
Q.62 A value of θ for which θ−θ+
sini21sini32 is purely imaginary, is :
(1) 3π (2)
6π (3) sin–1
⎟⎟⎠
⎞⎜⎜⎝
⎛
43 (4) sin–1
⎟⎟⎠
⎞⎜⎜⎝
⎛
31
Students may find similar question in CP exercise sheet : [JEE Main, Chapter : Complex Number, Exercise # 2, Q.8] [JEE Advance, Chapter : Complex Number, Exercise # 1, Q.21]
Ans. [4] Sol. For purely imaginary, real part of the given complex number should be zero.
∴ Re ⎥⎦⎤
⎢⎣⎡
θ−θ+
sini21sini32 = 0
CAREER POINT
CAREER POINT, CP Tower, IPIA, Road No.1, Kota (Raj.), : 0744-6630500 | www.ecareerpoint.com Email: info@cpil.in Page # 34
[ CODE – E ]
Re ⎥⎦⎤
⎢⎣⎡
θ+θ+
×θ−θ+
sini21sini21
sini21sini32 = 0
Re ⎥⎦
⎤⎢⎣
⎡
θ+θ
+θ+θ−
22
2
sin41sin7i
sin41sin62 = 0
⇒ θ+θ−
2
2
sin41sin62 = 0
⇒ sin2θ = 31
⇒ sin θ = ± 3
1
⇒ θ = sin–1⎟⎟⎠
⎞⎜⎜⎝
⎛
31 , sin–1
⎟⎟⎠
⎞⎜⎜⎝
⎛−
31
Q.63 The sum of all real values of x satisfying the equation ( ) 60x4x22
5x5x−+
+− = 1 is
(1) 3 (2) –4 (3) 6 (4) 5
Students may find similar question in CP exercise sheet : [JEE Advance, Chapter : Log & Modulus , Exercise # 3, Q.14]
Ans. [1]
Sol. Given equation ( ) 60x4x22
5x5x−+
+− = 1 Case I : (1)Any finite real number = 1 ∴ (x2 – 5x + 5) = 1 and (x2 + 4x – 60) = any finite real number ⇒ (x2 – 5x + 4) = 0 ⇒ (x – 1)(x – 4) = 0 x = 1, 4 ⇒ x = 1, 4 satisfies given equation Case II : (–1)even integer = 1 ∴ (x2 – 5x + 5) = –1 and x2 + 4x – 60 = even integer ⇒ x2 – 5x + 6 = 0 ⇒ x = 2, 3 Check (A) : At x = 2 ⇒ x2 + 4x – 60 = –48 = (even integer)
⇒ x = 2 is root Check (B) : At x = 3 ⇒ x2 + 4x – 60 = –39 ≠ even integer
⇒ x = 3 is not root Case III : (Any non zero real number)0 = 1 ∴ (x2 – 5x + 5) is any non zero real number and x2 + 4x – 60 = 0 ⇒ (x + 10) (x – 6) = 0 ⇒ x = 6, –10 are the roots
∴ overall roots are 1, 4, 2, –10, 6
Hence sum of roots = 1 + 4 + 2 + 6 – 10 = 3
CAREER POINT
CAREER POINT, CP Tower, IPIA, Road No.1, Kota (Raj.), : 0744-6630500 | www.ecareerpoint.com Email: info@cpil.in Page # 35
[ CODE – E ]
Q.64 If A = ⎥⎦
⎤⎢⎣
⎡ −23ba5
and A adj A = AAT, then 5a + b is equal to
(1) –1 (2) 5 (3) 4 (4) 13
Students may find similar question in CP exercise sheet : [JEE Advance, Chapter : Matrix, Exercise # 5A, Q.28]
Ans. [2]
Sol. A = ⎥⎦
⎤⎢⎣
⎡ −23ba5
A.Adj A = A.AT (given)
⇒ |A| (I2) = ⎥⎦
⎤⎢⎣
⎡ −23ba5
· ⎥⎦
⎤⎢⎣
⎡− 2b
3a5 , {using property A.Adj A = |A| · I2}
⇒ (10a + 3b) · ⎥⎦
⎤⎢⎣
⎡1001
= ⎥⎦
⎤⎢⎣
⎡
−−+
13b2a15b2a15ba25 22
, {|A| = 10a + 3b}
⇒ ⎥⎦
⎤⎢⎣
⎡+
+b3a100
0b3a10 = ⎥
⎦
⎤⎢⎣
⎡
−−+
13b2a15b2a15ba25 22
⇒ 15a – 2b = 0 ...(1)
⇒ 10a + 3b = 13 ...(2)
⇒ 25a2 + b2 = 13 ...(3) By solving eq. (1) and (2)
a = 52 ; b = 3
Check eq. (3)
Also At a = 52 , b = 3
25a2 + b2 = ⎟⎠⎞
⎜⎝⎛ ×
25425 + 9
= 13 satisfying equation (3)
∴ 5a + b = ⎟⎠⎞
⎜⎝⎛ ×
525 + 3 = 2 + 3 = 5
Q.65 The system of linear equations x + λy – z = 0 λx – y – z = 0 x + y – λz = 0 has a non-trivial solution for
(1) infinitely many values of λ (2) exactly one value of λ (3) exactly two values of λ (4) exactly three values of λ
CAREER POINT
CAREER POINT, CP Tower, IPIA, Road No.1, Kota (Raj.), : 0744-6630500 | www.ecareerpoint.com Email: info@cpil.in Page # 36
[ CODE – E ]
Students may find similar question in CP exercise sheet : [JEE Main, Chapter : Determinant, Exercise # 1, Q.41] [JEE Advance, Chapter : Determinant, Exercise # 3, Q.3]
Ans. [4] Sol. For non-trivial solution, Δ = 0
Δ = λ−
−−λ−λ
111111
= 0
⇒ 1(λ + 1) – λ(–λ2 + 1) – 1(λ + 1) = 0
⇒ λ3 – λ = 0
⇒ λ(λ2 – 1) = 0
⇒ λ = 0, 1, –1
∴ exactly three values of λ exist for which the given system of linear equations has non-trivial solution.
Q.66 If all the words (with or without meaning) having five letters, formed using the letters of the word SMALL and arranged as in a dictionary; then the position of the word SMALL is :
(1) 46th (2) 59th (3) 52nd (4) 58th
Students may find similar question in CP exercise sheet : [JEE Main, Chapter :P & C, Exercise # 5A, Q.8] [JEE Advance, Chapter : P & C, Exercise # 5B, Q.3]
Ans. [4] Sol. Method-I: Short Trick
LLAMS
SMLLA
//
//
S M A L L
)!01(!2
)!10(!2
)!20(!2
)!33(!2
)!44(×+
×+
×+
×+
×
= 48 + 9 + 0 + 0 + 1 = 58th rank S
MLLA
//
//
Method -II : Actual Method
(1) Words start with A ⇒ A
S, M , L, L
— — — — = !2!4 = 12
CAREER POINT
CAREER POINT, CP Tower, IPIA, Road No.1, Kota (Raj.), : 0744-6630500 | www.ecareerpoint.com Email: info@cpil.in Page # 37
[ CODE – E ]
(2) Words start with L ⇒ L
A, S , L, M
— — — — = 4! = 24
(3) Words start with M ⇒ M
A, S , L, L
— — — — = !2!4 = 12
(4) Words start with SA ⇒
A
M , L, L
— — — S = !2!3 = 3
(5) Words start with SL ⇒
L
A , M, L
— — — S = 3! = 6
(6) Words start with SMALL ⇒ LLAMS = 1
Total rank = 12 + 24 + 12 +3 + 6 +1 = 58
Q.67 If the number of terms in the expansion of n
2x4
x21 ⎟
⎠⎞
⎜⎝⎛ +− , x ≠ 0, is 28, then the sum of the coefficients of all the
terms in this expansion, is (1) 64 (2) 2187 (3) 243 (4) 729
Students may find similar question in CP exercise sheet : [JEE Advance, Chapter : Binomial Theorem, Exercise # 1, Q.23]
Ans. [Bonus] Sol. Given expansion
n
2x4
x21 ⎟
⎠⎞
⎜⎝⎛ +− , x ≠ 0,
n2
n2
x)4x2x( +− , x ≠ 0
No. of terms in given polynomial = 2n + 1 = 28
n = 2
27 which is not possible
If multinomial theorem is used then n
2x4
x21 ⎟
⎠⎞
⎜⎝⎛ +− = n2
n2
x)4x2x( +−
then no. of terms = n + 3 – 1C3–1 = 28
⇒ n+2C2 = 28 ⇒ !2
)1n)(2n( ++ = 28
⇒ (n + 2) (n + 1) = 8 × 7
CAREER POINT
CAREER POINT, CP Tower, IPIA, Road No.1, Kota (Raj.), : 0744-6630500 | www.ecareerpoint.com Email: info@cpil.in Page # 38
[ CODE – E ]
⇒ n + 2 = 8
⇒ n = 6
∴ sum of coefficient (put x = 1) = 6n
14
121
=
⎟⎠⎞
⎜⎝⎛ +− = (3)6 = 729
Although, application of multinomial theorem for one variable expression is conceptually wrong, hence bonus must be awarded.
Q.68 If the 2nd, 5th and 9th terms of a non-constant A.P. are in G.P., then the common ratio of this G.P. is
(1) 58 (2)
34 (3) 1 (4)
47
Students may find similar question in CP exercise sheet :
[JEE Main, Chapter : Progression, Exercise # 1, Q.41]
[JEE Advance, Chapter : Progression, Exercise # 3, Q. 4]
Ans. [2]
Sol. Let a be the first term and d is the common difference of A.P.
Now 2nd, 5th and 9th terms of A.P. are
T2 = a + d
T5 = a + 4d
T9 = a + 8d
These terms are in G.P.
∴ (a + 4d)2 = (a + d) (a + 8d)
⇒ a2 + 16d2 + 8ad = a2 + 9ad + 8d2
⇒ 8d2 = ad
⇒ 8d = a (d ≠ 0)
Now common ratio r = dad4a
++ =
dd8d4d8
++ =
34
Q.69 If the sum of the first ten terms of the series 2
2222
5444
513
522
531 ⎟
⎠⎞
⎜⎝⎛++⎟
⎠⎞
⎜⎝⎛+⎟
⎠⎞
⎜⎝⎛+⎟
⎠⎞
⎜⎝⎛ + ...., is
516 m, then m is
equal to
(1) 102 (2) 101 (3) 100 (4) 99
Students may find similar question in CP exercise sheet :
[JEE Advance, Chapter : Progression, Exercise # 5A, Q.24]
Ans. [2]
Sol. 251 [82 + 122 + 162 + 202 + ...... + upto 10 terms] =
516 m
CAREER POINT
CAREER POINT, CP Tower, IPIA, Road No.1, Kota (Raj.), : 0744-6630500 | www.ecareerpoint.com Email: info@cpil.in Page # 39
[ CODE – E ]
2
2
54 [22 + 33 + .... + 112] =
516 m
51 [(12 + 22 + ..... + 112) – 12] = m
⎥⎦
⎤⎢⎣
⎡⎟⎠⎞
⎜⎝⎛ −
×× 16
23121151 = m
(22 × 23 – 1) = 5m m = 101
Q.70 Let p = x21
2
0x)xtan1(lim +
+→ then log p is equal to
(1) 2 (2) 1 (3) 21 (4)
41
Ans. [3]
Sol. Indeterminate form 1∞.
So p = x21).x(tanlim 2
0xe+→
⇒ p = e1/2
⇒ log p = 21
Q.71 For x ∈ R, f(x) = |log 2 – sinx| and g(x) = f(f(x)), then :
(1) g is not differentiable at x = 0
(2) g′(0) = cos(log 2)
(3) g′(0) = – cos(log 2)
(4) g is differentiable at x = 0 and g′(0) = – sin(log2)
Students may find similar question in CP exercise sheet :
[JEE Advance, Chapter :AOD, Exercise # 5A, Q.9]
Ans. [2]
Sol. f(x) = log 2 – sin x, { ∵ when x → 0, log 2 – sinx > 0 }
g(x) = log 2 – sin (log 2 – sin x)
⇒ g′(x) = 0 + cos (log 2 – sin x) . cos x
⇒ g′(0) = cos(log 2)
CAREER POINT
CAREER POINT, CP Tower, IPIA, Road No.1, Kota (Raj.), : 0744-6630500 | www.ecareerpoint.com Email: info@cpil.in Page # 40
[ CODE – E ]
Q.72 Consider f(x) = tan–1⎟⎟⎠
⎞⎜⎜⎝
⎛
−+
xsin1xsin1 , x ∈ ⎟
⎠⎞
⎜⎝⎛ π
2,0 . A normal to y = f(x) at x =
6π also passes through the point :
(1) (0, 0) (2) ⎟⎠⎞
⎜⎝⎛ π
32,0 (3) ⎟
⎠⎞
⎜⎝⎛ π 0,
6 (4) ⎟
⎠⎞
⎜⎝⎛ π 0,
4
Students may find similar question in CP exercise sheet : [JEE Main, Chapter : AOD-1, Exercise # 5A, Q.31] [JEE Advance, Chapter : AOD-1, Exercise # 1, Q. 5]
Ans. [2]
Sol. f(x) = tan–1
xsin1xsin1
−+ , x ∈ ⎟
⎠⎞
⎜⎝⎛ π
2,0
f(x) = tan–1
2xsin–
2xcos
2xsin
2xcos +
, 2x ∈ ⎟
⎠⎞
⎜⎝⎛ π
4,0
⇒ f(x) = tan–1
⎟⎟⎟⎟
⎠
⎞
⎜⎜⎜⎜
⎝
⎛ +
2xsin–
2xcos
2xsin
2xcos
⇒ f(x) = tan–1 tan ⎟⎠⎞
⎜⎝⎛ +
π2x
4
⇒ f(x) = 4π +
2x
⇒ f ′(x) = 21
⇒ Slope of tangent = 21
⇒ Slope of normal = –2
∴ Equation of normal at ⎟⎠⎞
⎜⎝⎛ ππ
3,
6is
y – 3π = –2 ⎟
⎠⎞
⎜⎝⎛ π
6–x
⇒ y = – 2x +3
2π passes through ⎟⎠⎞
⎜⎝⎛ π
32,0
CAREER POINT
CAREER POINT, CP Tower, IPIA, Road No.1, Kota (Raj.), : 0744-6630500 | www.ecareerpoint.com Email: info@cpil.in Page # 41
[ CODE – E ]
Q.73 A wire of length 2 units is cut into two parts which are bent respectively to form a square of side = x units and a circle of radius = r units. If the sum of the areas of the square and the circle so formed is minimum, then :
(1) 2x = (π + 4)r (2) (4 – π)x = πr (3) x = 2r (4) 2x = r
Students may find similar question in CP exercise sheet : [JEE Main, Chapter : AOD-3 (max.-min.), Exercise # 1, Q.68] [JEE Advance, Chapter : AOD-3 (max.-min.), Exercise # Board pattern, Q.17]
Ans. [3] Sol.
Circle
O P
Sq. (side = x units)(Radius = r) units
Q
OQ + QP = 2 (given)
⇒ 4x + 2πr = 2
⇒ 2x + πr = 1 … (1)
r = π
x2–1
Sum of areas = A = x2 + πr2
A = x2 +π 2
2)x2–1(π
dxdA = 2x +
π)x2–1(2 (–2)
⇒ 2x – 4r = 0
⇒ 4r = 2x
⇒ 2r = x
Q.74 The integral ∫ +++
335
912
)1xx(x5x2 dx is equal to -
(1) 235
5
)1xx(x–
+++C (2) 235
10
)1xx(2x
+++C
(3) 235
5
)1xx(2x
+++C (4) 235
10
)1xx(2x–
+++C
where C is an arbitrary constant.
Students may find similar question in CP exercise sheet : [JEE Main, Chapter :Ind. Integral, Exercise # 4, Q.19]
Ans. [2]
Sol. I = ∫ +++
335
912
)1xx(x5x2 dx
CAREER POINT
CAREER POINT, CP Tower, IPIA, Road No.1, Kota (Raj.), : 0744-6630500 | www.ecareerpoint.com Email: info@cpil.in Page # 42
[ CODE – E ]
I = ∫⎟⎠⎞
⎜⎝⎛ ++
+
3
52
63
x1
x11
x5
x2
dx
Let 1 + 2x
1 + 5x1 = t
⇒ ⎟⎠⎞
⎜⎝⎛
63 x5–
x2– dx = dt
∴ I = ∫ 3tdt–
⇒ I = – ∫ dtt 3–
⇒ I = 2–
t– 2–+ C = 2t2
1 +C
⇒ I = 2
52 x1
x112
1
⎟⎠⎞
⎜⎝⎛ ++
+ C
⇒ I = 235
10
)1xx(2x
++ + C
Q.75 ∞→n
limn/1
n2nn3)....2n)(1n(
⎟⎠⎞
⎜⎝⎛ ++ is equal to -
(1) 4e18 (2) 2e
27 (3) 2e9 (4) 3 log3 – 2
Students may find similar question in CP exercise sheet : [JEE Main, Chapter :Def. Integral , Exercise # 5B, Q.27] [JEE Advance, Chapter : Def. Integral, Exercise # 4,Q.26]
Ans. [2]
Sol. y = n/1
nn2n.....
n2n
n1n
⎥⎦
⎤⎢⎣
⎡⎟⎠⎞
⎜⎝⎛ +
⎟⎠⎞
⎜⎝⎛ +
⎟⎠⎞
⎜⎝⎛ +
Taking log both sides
log y = ∑=
⎟⎠⎞
⎜⎝⎛ +
n2
1r nr1log
n1
⇒ log y = ∫ +2
0
dx)x1log(
Put 1 + x = t ∴ dx = dt
CAREER POINT
CAREER POINT, CP Tower, IPIA, Road No.1, Kota (Raj.), : 0744-6630500 | www.ecareerpoint.com Email: info@cpil.in Page # 43
[ CODE – E ]
⇒ log y = ∫3
1
dttlog = 31t–tlogt
⇒ log y = 3 log 3 – 3 – [–1]
⇒ log y = log 27 – 2
⇒ y = elog 27 – 2
⇒ y = 27e–2 Q.76 The area (in sq. units) of the region {(x, y) : y2 ≥ 2x and x2 + y2 ≤ 4x, x ≥ 0, y ≥ 0} is -
(1) π –34 (2) π –
38 (3) π –
324 (4)
2π –
322
Students may find similar question in CP exercise sheet : [JEE Advance, Chapter : Area under the curve, Exercise # 5A, Q.2]
Ans. [2] Sol.
P(2, 2)
(2, 0)
Q(2, –2)
(0, 0) X
Y
⇒ y2 ≥ 2x ...(1)
⇒ x2 + y2 ≤ 4x
⇒ x2 – 4x + y2 ≤ 0
⇒ (x – 2)2 + y2 ≤ 22 ...(2) for point of intersection x2 + 2x = 4x x2 – 2x = 0 x(x – 2) = 0 x = 0, 2
∴ at x = 0, y = 0 and at x = 2, y = ± 2
∴ P(2, 2) and Q(2, –2) Hence required area
= ∫2
0
2 dx)x2–x–x4(
= ∫2
0
2 dx)x2–)2–x(–4(
CAREER POINT
CAREER POINT, CP Tower, IPIA, Road No.1, Kota (Raj.), : 0744-6630500 | www.ecareerpoint.com Email: info@cpil.in Page # 44
[ CODE – E ]
= 2
0
2/31–2
2/3x2–
22–xsin
24x–x4
22–x
⎥⎥⎦
⎤
⎢⎢⎣
⎡⎟⎠⎞
⎜⎝⎛+
= ⎥⎦
⎤⎢⎣
⎡⎟⎠⎞
⎜⎝⎛ π
××2
–24–
3222– 2/3
= ⎥⎦⎤
⎢⎣⎡ π+
×3
24– = π – 38
Q.77 If a curve y = f(x) passes through the point (1, –1) and satisfies the differential equation, y(1 + xy)dx = x dy, then
f ⎟⎠⎞
⎜⎝⎛
21– is equal to -
(1) –52 (2) –
54 (3)
52 (4)
54
Students may find similar question in CP exercise sheet : [JEE Advance, Chapter : Differential Equation, Exercise # 3, Page No. 3, Q.13]
Ans. [4]
Sol. y(1 + xy) = xdxdy
dxdy =
xy + y2
⇒ dxdy –
xy = y2
⇒ 2y1
dxdy –
x1 .
y1 = 1 Let –
y1 = t ⇒ 2y
1dxdy =
dxdt
dxdt +
x1 .t = 1
I.F. = ∫ dx
x1
e = e n x
I.F. = x Solution of this differential equation is
⇒ t.x = ∫ dxx
⇒ tx = 2
x2+ C
⇒ –yx =
2x2
+ C this curve passes through (1, –1) so
– (–1) = 21 + C ⇒ C =
21
CAREER POINT
CAREER POINT, CP Tower, IPIA, Road No.1, Kota (Raj.), : 0744-6630500 | www.ecareerpoint.com Email: info@cpil.in Page # 45
[ CODE – E ]
⇒ – yx =
2x2
+21 ....(1)
from at x = –21
y21 =
81 +
21 ⇒
y1 =
41 + 1 ⇒ y =
54
Q.78 Two sides of a rhombus are along the lines, x – y + 1 = 0 and 7x – y – 5 = 0. If its diagonals intersect at (–1, –2),
then which one of the following is a vertex of this rhombus ?
(1) (–3, –9) (2) (–3, –8) (3) ⎟⎠⎞
⎜⎝⎛
38–,
31 (4) ⎟
⎠⎞
⎜⎝⎛
37–,
310–
Ans. [3] Sol.
P
B
C (x1, y1)D
A Equation of AB ⇒ x – y + 1 = 0 … (1)
Equation of AD ⇒ 7x – y – 5 = 0 … (2)
Solve (1) & (2) ⇒ A(1, 2)
∵ Point P(–1, –2) is the mid-point of the diagonals AC & BD
∵ AP = CP ⇒ C ⇒ ⎪⎩
⎪⎨
⎧
=⇒=+
=⇒=+
6–y2–2
2y
3–x1–2
1x
11
11
C(–3, –6)
∴ Equation of BC ⇒ 7x – y + λ1 = 0 It's passes through the point C, then ⇒ 7(–3) – (–6) + λ1 = 0
⇒ λ1 = 15
∴ Equation of BC is ⇒ 7x – y + 15 = 0 … (3)
Solve (1) & (3) ⇒ B ⎟⎠⎞
⎜⎝⎛
34–,
37–
∴ Equation of CD ⇒ x – y + λ2 = 0
⇒ x – y – 3 = 0 … (4)
Solve (2) & (4) ⇒ D ⎟⎠⎞
⎜⎝⎛
38–,
31
CAREER POINT
CAREER POINT, CP Tower, IPIA, Road No.1, Kota (Raj.), : 0744-6630500 | www.ecareerpoint.com Email: info@cpil.in Page # 46
[ CODE – E ]
Q.79 The centres of those circles which touch the circle, x2 + y2 – 8x – 8y – 4 = 0, externally and also touch the x-axis, lie on :
(1) a circle (2) an ellipse which is not a circle (3) a hyperbola (4) a parabola
Students may find similar question in CP exercise sheet : [JEE Main, Chapter :Circle, Exercise # 4, Q.81] [JEE Advance, Chapter : Parabola, Exercise # 5B, , Q.7]
Ans. [4] Sol. x2 + y2 – 8x – 8y – 4 = 0 … (1) C1 = centre (4, 4)
r1 = radius = 41616 ++ = 6
(1)
C1(4, 4)(2)
C2
xMO (0, 0)
y
r1 r2
Locus of the centre C2(h, k) = ?
∵ radius of circle (2) is
r2 = | k | & circle (2) touches the circle (1) externally, So, C1C2 = r1 + r2
⇒ (h + 4)2 + (k – 4)2 = (6 + | k |)2
⇒ h2 + 16 – 8h + k2 + 16 – 8k = 36 + k2 + 12| k |
⇒ Replaced h by x & k by y then
⇒ x2 – 8x – 8y – 12 | y | – 4 = 0
∵ | y | ⇒ ⎪⎩
⎪⎨⎧
<
≥
0yIf;y–or
0yIf;y
Case (I) : If y ≥ 0
⇒ x2 – 8x – 20y – 4 = 0 Case (II) : If y < 0
⇒ x2 – 8x + 4y – 4 = 0 Both of these equations are the equation of parabola.
CAREER POINT
CAREER POINT, CP Tower, IPIA, Road No.1, Kota (Raj.), : 0744-6630500 | www.ecareerpoint.com Email: info@cpil.in Page # 47
[ CODE – E ]
Q.80 If one of the diameters of the circle, given by the equation, x2 + y2 – 4x + 6y – 12 = 0, is a chord of a circle S, whose centre is at (–3, 2), then the radius of S is -
(1) 25 (2) 35 (3) 5 (4) 10
Students may find similar question in CP exercise sheet : [JEE Main, Chapter :Circle, Exercise # 5A, Q.13] [JEE Advance, Chapter : Circle, Exercise # 5A, Q.18]
Ans. [2] Sol.
C2 C1 (1)
B
A
(2)
Diameter of circle (1)
x2 + y2 – 4x + 6y – 12 = 0 … (1) centre C1 (2, –3) radius r1 = 5 & centre of circle (2) is C2 (–3, 2)
∴ C1C2 = 22 )2–3(–)32( ++ = 2525 + = 25
∴ radius (r2) of circle (2) is
⇒ r2 = C2A = 221
21 CCAC + = 22 )25(5 +
= 5025 + = 75 = 35
Q.81 Let P be the point on the parabola, y2 = 8x which is at a minimum distance from the centre C of the circle,
x2 + (y + 6)2 = 1. Then the equation of the circle, passing through C and having its centre at P is - (1) x2 + y2 – 4x + 8y + 12 = 0 (2) x2 + y2 – x + 4y – 12 = 0
(3) x2 + y2 – 4x + 2y – 24 = 0 (4) x2 + y2 – 4x + 9y + 18 = 0
Ans. [1] Sol. y2 = 8x Let, the point P (2t2, 4t) lies on the parabola & centre of the given circle is C (0, –6) ∴ Distance between the points C & P.
2)6t4(44D +++=
For minimum distance
⇒ 0t4t42
4)6t4(2)34(4dtdD
4=
+
×+++=
CAREER POINT
CAREER POINT, CP Tower, IPIA, Road No.1, Kota (Raj.), : 0744-6630500 | www.ecareerpoint.com Email: info@cpil.in Page # 48
[ CODE – E ]
4t3 + 8t + 12 = 0 t3 + 2t + 3 = 0 t2(t + 1) – t(t + 1) + 3(t + 1) = 0 (t2 – t + 3) (t + 1) = 0 t = –1 ∴ point P (2, –4) centre C (0, –6)
∴ CP = radius of circle = 2244 =+
∴ Equation of circle ⇒ (x – 2)2 + (y + 4)2 = 8 x2 – 4x + 4 + y2 + 8y + 16 = 8 x2 + y2 – 4x + 8y + 12 = 10 Q.82 The eccentricity of the hyperbola whose length of the latus rectum is equal to 8 and the length of its conjugate axis
is equal to half of the distance between its foci, is -
(1) 34 (2)
34 (3)
32 (4) 3
Students may find similar question in CP exercise sheet : [JEE Main, Chapter : Hyperbola, Exercise # 1, Q.1]
Ans. [3]
Sol. Let eqn. of hyperbola is 1by
ax
2
2
2
2=−
It’s length of latus-rectum is 8ab2 2
=
⇒ b2 = 4a …(1) & length of it’s conjugate axis = half of the distance between it’s foci
⇒ 21b2 = (2ae)
⇒ 2aeb = …(2)
∵ b2 = a2(e2 – 1) …(3)
By eqn (2) & (3)
⇒ )1e(a4ea 22
22−=
⇒ 14ee
22 =−
⇒ 14e3 2
=
⇒ 3
2e =
CAREER POINT
CAREER POINT, CP Tower, IPIA, Road No.1, Kota (Raj.), : 0744-6630500 | www.ecareerpoint.com Email: info@cpil.in Page # 49
[ CODE – E ]
Q.83 The distance of the point (1, –5, 9) from the plane x – y + z = 5 measured along the line x = y = z is :
(1) 103 (2) 310 (3) 3
10 (4) 3
20
Students may find similar question in CP exercise sheet : [JEE Main, Chapter :3D, Exercise #3, Q.18]
Ans. [2] Sol. Given plane x – y + z = 5 …(1) and line x = y = z …(2)
P(1,–5,9)
Plane 1
M
Line 2
∵ Given line is parallel to line PM
∴ DR’s of line PM = DR’s of given line = 1, 1, 1
∴ Eqn. of line PM is ⇒ λ=−
=+
=−
19z
15y
11x (Let)
∴ Coordinates of point M is (λ + 1, λ – 5, λ + 9)
∵ Point M lies on the plane x – y + z = 5
That’s why ⇒ (λ + 1) – (λ – 5) + (λ + 9) = 5
⇒ λ + 15 = 5
λ = –10
∴ M(–9, –15, –1)
∴ 100100100)19()155()91(PM 222 ++=+++−++=
310=
Q.84 If the line, 3
4z12y
23x +
=−+
=− lies in the plane, lx + my – z = 9, then l2 + m2 is equal to :
(1) 26 (2) 18 (3) 5 (4) 2
Students may find similar question in CP exercise sheet : [JEE Main, Chapter :3D, Exercise #4, Q.20] [JEE Advance, Chapter : 3D, Exercise # 3, Q.21]
Ans. [4]
CAREER POINT
CAREER POINT, CP Tower, IPIA, Road No.1, Kota (Raj.), : 0744-6630500 | www.ecareerpoint.com Email: info@cpil.in Page # 50
[ CODE – E ]
Sol.
PLine 1
Plane x + my – z = 9
34z
12y
23x +
=−+
=− ...(1) lie on plane
Let point P lies on line (1)
∴ P(3, –2, –4)
∴ Point P also lies on plane lx + my – z = 9 ....(2)
3l – 2m + 4 = 9 ⇒ 3l – 2m = 5 …(3)
2l – m – 3 = 0 ⇒ 4l – 2m = 6 …(4) On solving eq. (3) & (4) – + – – l = – 1
⇒ l = 1 m = –1
∴ l2 + m2 = 2
Q.85 Let b,a and c be three unit vectors such that .)cb(23)cb(a ×=×× If b is not parallel to c , then the angle
between a and b is :
(1) 4
3π (2) 2π (3)
32π (4)
65π
Students may find similar question in CP exercise sheet : [JEE Main, Chapter : Vector, Exercise # 5A, Q.4] [JEE Advance, Chapter : Vector, Exercise # 1, Q.29]
Ans. [4]
Sol. )cb(23)cb(a +=××
c23b
23c)ba(b)ca( +=⋅−⋅
Now comparing both sides ( b and c are not parallel)
⇒ 23ca =⋅ &
23ba −=⋅
CAREER POINT
CAREER POINT, CP Tower, IPIA, Road No.1, Kota (Raj.), : 0744-6630500 | www.ecareerpoint.com Email: info@cpil.in Page # 51
[ CODE – E ]
∵ | a | = | b | = 1 and if angle between a and b is θ then
23cos −=θ
6
5π=θ
Q.86 If the standard deviation of the numbers 2, 3, a and 11 is 3.5, then which of the following is true ?
(1) 3a2 – 26a + 55 = 0 (2) 3a2 – 32a + 84 = 0 (3) 3a2 – 34a + 91 = 0 (4) 3a2 – 23a + 44 = 0
Ans. [2] Sol. Observations are 2, 3, a, 11
Mean 4a4
411a32x +=
+++= and σ2 = (3.5)2 = 12.25
Now, σ2 = n1 2
n
1i
2i )x(x −∑
=
2
2
4a4]121a94[
4125.12 ⎟
⎠⎞
⎜⎝⎛ +−+++=
⇒ a216a16
4a13425.12
22−−−
+=
⇒ a216a16
4a5.3325.12
22−−−+=
⇒ 025.5a216a3 2
=+−
⇒ 3a2 – 32a + 84 = 0 Q.87 Let two fair six-faced dice A and B be thrown simultaneously. If E1 is the event that die A shows up four, E2 is
the event that die B shows up two and E3 is the event that the sum of numbers on both dice is odd, then which of the following statements is NOT true ?
(1) E1 and E2 are independent (2) E2 and E3 are independent (3) E1 and E3 are independent (4) E1, E2 and E3 are independent
Students may find similar question in CP exercise sheet : [JEE Advance, Chapter : Probability, Exercise # 4, Q.27]
Ans. [4]
Sol. P(E1) = 61 E1 = shows 4
P(E2) = 61 E2 = shows 2
CAREER POINT
CAREER POINT, CP Tower, IPIA, Road No.1, Kota (Raj.), : 0744-6630500 | www.ecareerpoint.com Email: info@cpil.in Page # 52
[ CODE – E ]
P(E3) = 3618 E3 = sum equal to odd on both dice
P(E1 ∩ E2) = 361 E1 ∩ E2 = )}2,4{(
P(E1 ∩ E3) 363
= E1 ∩ E3 = {(4, 1), (4, 3), (4, 5)}
P(E2 ∩ E3) 363
= E2 ∩ E3 = {(2, 1), (2, 3), (2, 5)}
E1 ∩ E2 ∩ E3 = { } Check by options
(1) P(E1 ∩ E2) = P(E1) × (E2)
61
61
361
×= ⇒ E1 and E2 are independent → True
(2) P(E1 ∩ E2) = P(E1) × P(E2)
61
61
361
×= ⇒ E1 and E2 are independent → True
(3) P(E1 ∩ E3) = P(E1) × P(E3)
3618
61
363
×= ⇒ E1 and E3 are independent → True
(4) P(E1 ∩ E2 ∩ E3) ≠ P(E1) × P(E2) × P(E3)
0 ≠ 3618
61
61
×× ⇒ E1, E2, E3 are not independent → False
Q.88 If 0 ≤ x < 2π, then the number of real values of x, which satisfy the equation cos x + cos 2x + cos 3x + cos 4x = 0, is :
(1) 3 (2) 5 (3) 7 (4) 9
Students may find similar question in CP exercise sheet : [JEE Advance, Chapter : Trigonometric equation, Exercise # 1,Q.12]
Ans. [3] Sol. cos x + cos 2x + cos 3x + cos 4x = 0
⇒ 0
2xsin
2x4sinx
214xcos
=⎟⎠⎞
⎜⎝⎛⋅⎥
⎦
⎤⎢⎣
⎡⎟⎠⎞
⎜⎝⎛ −
+ [By using series formula]
⇒ cos 0x2sin2x5
=⋅ But ⎟⎠⎞
⎜⎝⎛ ≠ 0
2xsin
∴ sin 2x = 0 cos 02x5
=
⇒ 2x = nπ ⇒ 2
)1n2(2x5 π
+=
CAREER POINT
CAREER POINT, CP Tower, IPIA, Road No.1, Kota (Raj.), : 0744-6630500 | www.ecareerpoint.com Email: info@cpil.in Page # 53
[ CODE – E ]
⇒ In,2
nx ∈π
= ⇒ x = (2n + 1) ,5π n ∈ I
⇒ x = 0, ,2π π, ,
23π 2π
59,
57,,
53,
5x ππ
πππ
=
But x = 0, 2π does not satisfy (∵ sin2x ≠ 0)
∴ Solutions are 5
9,5
7,5
3,5
,2
3,,2
x ππππππ
π=
7 solutions Q.89 A man is walking towards a vertical pillar in a straight path, at a uniform speed. At a certain point A on the
path, he observes that the angle of elevation of the top of the pillar is 30°. After walking for 10 minutes from A in the same direction, at a point B, he observes that the angle of elevation of the top of the pillar is 60°. Then the time taken (in minutes) by him, from B to reach the pillar, is :
(1) 6 (2) 10 (3) 20 (4) 5
Students may find similar question in CP exercise sheet : [JEE Main, Chapter : _Height and distance, Exercise # 1,Q.21] [JEE Advance, Chapter : Height and distance, Exercise # 4,Q.7]
Ans. [4] Sol.
30° 60° A B
y x
D
h
C
In ΔACD, yx
h30tan+
=°
⇒ 3yxh +
=
Now in ΔBCD, tan 60° yh
=
⇒ hy3 = ⇒ 3yxy3 +
= ⇒ x = 2y ⇒ y = 2x
Given that travelling time x = 10 min
∴ travelling time min52
10y ==
CAREER POINT
CAREER POINT, CP Tower, IPIA, Road No.1, Kota (Raj.), : 0744-6630500 | www.ecareerpoint.com Email: info@cpil.in Page # 54
[ CODE – E ]
Q.90 The Boolean Expression (p ∧ ~ q) ∨ q ∨ (~ p ∧ q) is equivalent to : (1) ~ p ∧ q (2) p ∧ q (3) p ∨ q (4) p ∨ ~ q
Ans. [3] Sol.
p q ~p ~q (p ∧ ~q) (1)
(~p ∧ q) (p ∧ ~q) ∨ q
(p ∧ ~q) ∨ q ∨ (~p ∧ q)
(2) p ∧ q
(3) p ∨ q
(4) (p ∨ ~q)
T T F F F F T T T T T T F F T T F T T F T T F T T F F T T T F T T F F T T F F F F F F T
top related